Given that $f(z)$ is an entire function and is not identically zero. Also $f(z)$ has no zeros on the real axis. Consider a small rectangle $R_{\epsilon(H)}=\{x+iy\in\mathbb{C}\mid 3-\epsilon(H)\leq x\leq 3+\epsilon(H),\ 0\leq y\leq H+\epsilon(H)\}$ where $H>4$ and $\epsilon(H)>0$ is a continuous function depending on $H$.
Question: Prove that there exists $\epsilon_0>0$ such that $f(z)$ has no zeros on the boundary of the rectangle $R_{\epsilon(H)}$ whenever $0<\epsilon(H)<\epsilon_0$.
My attempt: We prove the above question by the method of contradiction. If possible let us suppose no such $\epsilon_0$ exists. Then for each $\epsilon(H)>0$, there must exist $\epsilon'(H)$ with $0<\epsilon'(H)<\epsilon(H)$ such that the boundary of the rectangle $R_{\epsilon'(H)}$ has atleast one zero of $f$.
In particular if we take $\epsilon_n(H)=\epsilon^*(H)/2^n$ where $n\in \mathbb{N}$ and $0<\epsilon^*(H)\leq 1$, then there exists $\epsilon_{n}'(H)$ such that $0<\epsilon'_n(H)<\epsilon_n(H)=\epsilon^*(H)/2^n$ such that the boundary of $R_{\epsilon_n'(H)}$ has atleast one zero $z_n$ of $f$.
Without loss of generality we may assume that $\{\epsilon_n'(H)\}_{n\geq 1}$ is decreasing and $\epsilon_n'(H)\to 0^+$ (to see this choose $\epsilon_1'(H)<\epsilon_1(H)$ and then choose $\epsilon_2'(H)<\epsilon_1'(H)$ and $\epsilon_2'(H)<\epsilon^*(H)/2^2$ and we continue this process).
Now we consider a set $A=\{z_n\in \mathbb{C}\mid n\in \mathbb{N}\}$ where $z_n$ are those zeros of $f$ as we have chosen above. Now we can have two cases:
Case $1$: $A$ is an infinite set: Since $\epsilon_n(H)=\epsilon^*(H)/2^n$ so we have $\epsilon^*(H)>\epsilon_1(H)>\epsilon_2(H)>\epsilon_3(H)>...$ and hence all the zeros $z_n$ of set $A$ lie inside the rectangle $R_{\epsilon^*(H)}$ where $0<\epsilon^*(H)\leq 1$ and clearly $R_{\epsilon^*(H)}$ is a bounded set. Hence we get that $A$ is also a bounded set.
So by the Bolzano--Weierstrass theorem, set $A$ has a limit point. Since $f$ is entire so by the Identity theorem since $A\subset \mathbb{C}$ is an infinite bounded subset of the zeros of $f$ and it has a limit point so $f$ must be identically zero which is a contradiction.
Case $2$: $A$ is a finite set: In this case there are infinitely many $R_{\epsilon_n'(H)}$ having common zeros of $f$, thereby forcing zeros to lie on the real axis (as only the real axis can be common in $R_{\epsilon_n'(H)}$). This again contradicts the fact that $f$ has no zeros on the real axis.
I am thankful to you for helping me in this question.
It is not clear what the motivation for either question is, which makes this more difficult to answer. In any case, I do not believe that the first result is true.
I assume that $H \ge 0$ (so that $R_{\epsilon(H)} \neq \emptyset$). Then, consider the entire function:
$$f : \mathbb{C} \to \mathbb{C}, \;\;f(z)=\sinh(2z-3)$$
and $$\epsilon(H) = \lceil H \rceil \frac{\pi}{2} - H$$
where $\lceil x \rceil$ is the smallest integer greater than or equal to $x$, for $x \in \mathbb{R}$.
Now, note that for any choice of $H$, by construction, the boundary of $R_{\epsilon(H)}$ passes through the point $z = 3+i\lceil H \rceil\frac{\pi}{2}$, which is a zero for $f$, and it may readily be verified that for any $\epsilon_0>0$, there exists some $H$ such that $\epsilon_0/2 < \epsilon(H) < \epsilon_0$.
You might observe that my choice of $\epsilon : \mathbb{R} \to \mathbb{R}$ is not continuous. Maybe the result is true if it is, but without any motivation for the question, it is not clear that I should have assumed this. Similarly for $H > 4$, the fact that the rectangle is centred about $x = 3$, etc. -- it is completely unclear what relevance any of these constraints have to the question.
Please provide a clearer and better-motivated explanation for both questions.